Rates of Change — Question 7

Question 7 - Solution

Let:

• $x$: distance from person to wall (in m)

• $s$: length of the shadow on the wall (in m)

• Person’s height: 2 m

• Spotlight is 10 m from the wall

Using similar triangles (person is between wall and spotlight, so spotlight–person distance is $10-x$):

$$\frac{2}{10-x}=\frac{s}{10} \quad\Rightarrow\quad s=\frac{20}{10-x}$$

Differentiate with respect to $t$:

$$\frac{ds}{dt} = 20\cdot \frac{d}{dt}\left((10-x)^{-1}\right) = 20(10-x)^{-2}\cdot \frac{dx}{dt}$$

Substitute $x=6$ and $\frac{dx}{dt}=1.5$:

$$\frac{ds}{dt} = \frac{20(1.5)}{(10-6)^2} = \frac{30}{16} = \frac{15}{8}$$

Answer: $$\boxed{\frac{ds}{dt}=\frac{15}{8}\ \text{m/s}} \quad\text{(The shadow is increasing at this rate.)}$$